ME 314 Chapter 1 HW August 29, 2011

Chapter 1 HW Solution

Problem 1.1. Here are nine photos of planar four-bar linkages.

(a) Motorcycle shift linkage. (b) Garage door linkage. (c) Toolbox hinge linkage.

(d) Stepladder linkage. (e) 1968 GMC pickup hood linkage. (f) Card table folding leg linkage.

(g) Folding chair linkage. (h) Recliner chair linkage. (i) Briefcase hinge linkage.

Figure 1: Planar four-bar linkages in practice.

I have several photos of motorcycle engines in various stages of disassembly, showing portions of the valvetrain andcrankshaft/connecting rod linkages, but it’s difficult to see the entire linkage, so I left them out.

ME 314 Chapter 1 HW August 29, 2011

Problem 1.2. With link lengths of

link length

1 5 in2 1 in3 3 in4 5 in

this mechanism satisfies Grashof’s Law, and can continuously rotate. I’ll construct four ADAMS models of thislinkage, with consecutive links fixed, and animate them during class if time permits.

(a) Link 1 fixed. A screenshot of this mechanism is shown in Figure 1(a) below. Link 2 can rotate continuously,and this is a CRANK-ROCKER mechanism.

(b) Link 2 fixed. Link 2 is now the “ground,” and link 1 is the long blue link. Here link 4 (at the top) can rotatecontinuously, and this is a DRAG-LINK mechanism.

(c) Link 3 fixed. Link 2 can again rotate continuously, so this is another CRANK-ROCKER mechanism.

(d) Link 4 fixed. Link 2 is now the “ground,” and link 1 is the long blue link. Here link 4 (at the top) can rotatecontinuously, and this is a DOUBLE-ROCKER mechanism.

(a) Link 1 fixed: CRANK-ROCKER. (b) Link 2 fixed: DRAG-LINK.

(c) Link 3 fixed: CRANK-ROCKER. (d) Link 4 fixed: DOUBLE-ROCKER.

Figure 2: The four inversions of Problem 2.

ME 314 Chapter 1 HW August 29, 2011

Problem 1.4. I constructed the ADAMS model of this problem; I used two separate elements for link 3, the “merged”them. I selected a WHITE background in this case; may be easier to print. The path of point C is shown in black.

Figure 3: ADAMS model for Problem 4. Path of point C shown in black.

Problem 1.5: There are four mechanisms shown in Figure 4 (poorly-scanned figure on the next page); in each caseyou apply the Kutzbach criterion

m = 3(n− 1) − 2j1 − j2 (1)

in which

m = mobility (number of mechanism DOF)

n = number of links

j1 = number of single DOF joints

j2 = number of two DOF joints

ME 314 Chapter 1 HW August 29, 2011

Figure 4:

(a) Here n = 6, j1 = 7 (7 1DOF pin joints), and j2 = 0, so

m = 3(n− 1) − 2j1 − j2 = 3(5) − 2(7) − 0 = 1 DOF (I can visualize this)

(b) Here n = 8, j1 = 10 (10 1DOF pin joints), and j2 = 0, so

m = 3(n− 1) − 2j1 − j2 = 3(7) − 2(10) − 0 = 1 DOF (I can visualize this one, too)

(c) Here n = 7, j1 = 9 (8 1DOF pin joints plus rolling w/o slipping is 1 DOF), and j2 = 0, so

m = 3(n− 1) − 2j1 − j2 = 3(6) − 2(9) − 0 = 0 DOF (this is a little harder to see)

(d) Here n = 4, j1 = 3 (3 1DOF prismatic joints), and j2 = 2 (2 2DOF pin-in-slot joints), so

m = 3(n− 1) − 2j1 − j2 = 3(3) − 2(3) − 2 = 1 DOF (I definitely can’t visualize this)

ME 314 Chapter 1 HW August 29, 2011

Problem 1.16: REAL MODEL: I built this thing using cardboard and thumbtacks—see the picture below.

My wife and kids loved this model—there’s something about actually making a physical model that transcends sim-ulations. Anyway, the complete coupler curve of the Roberts mechanism is shown in the picture—I traced it using apen stuck through point P . Note that the segment between points A and B is quite straight.

By the way, from Grashof’s Law, 1.25 + 2.5 ≤ 2.5 + 2.5 so the coupler should have continuous rotation, and it does (ifyou maneuver the thumbtacks out of the way).

ADAMS MODEL (not required): I also created an ADAMS model of this mechanism and plotted the path ofpoint P :

They both look pretty close. But with the ADAMS model one can really examine the “straight-line” portion...

ME 314 Chapter 1 HW August 29, 2011

ADAMS plot of path of point P (not req’d). The plot below shows the same path, but plotted from within theADAMS Postprocessor. It’s the same path, although since the axes don’t have equal scaling it is a little “squeezed.”Between −1.25 < X < 1.25 inches there appears to be virtually no Y motion (this is the straight line).

43210-1-2-3-40.0

4.0

3.0

2.0

1.0

0.0

X displacement (inch)

Y di

spla

cem

ent (

inch

)

Figure 5: Path of Roberts mechanism point P from ADAMS Postprocessor.

Expanded plot of Y motion within straight line (not req’d.) Below is an expanded plot of one half the motionof point P between the center and the endpoint:

1.21.00.80.60.40.20.0

0.0

0.001

0.0

-0.001

-0.002

-0.003

-0.004

-0.005

X displacement (inch)

Y di

spla

cem

ent (

inch

)

Figure 6: Closeup of the straight-line segment showing 0.0028 inch Y error.

Inspection of the plot shows a maximum deviation of 0.0028 inches—less than three-thousandths of an inch. Prettygood straight line! Note the benefit of using ADAMS to simulate this mechanism.

ME 314 Chapter 1 HW August 29, 2011

Problem 1.17: Assume that both threads are right-handed. A thread pitch of 16NF means 16 threads/inch, and18NF is 18 threads/inch (the major diameters 3/4 and 5/8 inches are not relevant for this problem). Therefore, as thecrank is turned 10 revolutions clockwise, the entire screw will advance (move away from the crank) 10/16 inch, whilethe carriage will retreat (move towards the crank) 10/18 inch. The overall carriage motion is then:

δx =10

16− 10

18=

90 − 80

144=

5

72= 0.0694 in (2)

BTW—although not asked for—it is interesting to compute the effective “pitch” of the differential screw. Since 10revolutions result in 10/144 in displacement, one revolution would cause 1/144 in displacement. The effective pitchwould therefore be 144, equivalent to a single screw with 144 threads/in.

This is a pretty “fine” pitch, and would be difficult to obtain with a single thread.